随记-面试遇到的一些题

记录一些面试的题吧，没事自己看看

1.python,c++,java棋盘中的白子进行八连通区域查找，并返回连通区域块数和连通区域分类情况

# encoding=utf8


def show(L):
    # 直观显示用 非本题核心代码
    rows = len(L)
    cols = len(L[0])
    for i in range(rows):
        for j in range(cols):
            print(L[i][j], end=" ")
        print()


# 棋盘19*19 1为白色棋子  0为其他
L = [[0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1],
     [0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1],
     [0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1],
     [0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1]]

rows = len(L)
cols = len(L[0])
# 生成flag矩阵 标记处理过的点
flag = []
for i in range(rows):
    col = cols * [0]
    flag.append(col)


def means(i, j, k):
    ‘‘‘
    聚类查找
    :param i: 棋盘行坐标
    :param j: 棋盘列坐标
    :param k: 聚类类别
    ‘‘‘
    # 非白棋或者查找过的点排除
    if i < 0 or i >= rows or j < 0 or j >= cols or \
            flag[i][j] != 0 or L[i][j] != 1:
        return 0
    flag[i][j] = k
    # 使用八连通区域查找
    means(i, j - 1, k)
    means(i, j + 1, k)
    means(i - 1, j, k)
    means(i + 1, j, k)
    means(i + 1, j - 1, k)
    means(i + 1, j + 1, k)
    means(i - 1, j - 1, k)
    means(i - 1, j + 1, k)


if __name__ == "__main__":
    # k聚类
    k = 1
    for i in range(rows):
        for j in range(cols):
            # 未查找过的白棋
            if L[i][j] == 1 and flag[i][j] == 0:
                means(i, j, k)
                k += 1
    print(‘共有{}块‘.format(k))
    print(‘聚类效果如下：‘)
    show(flag)

2.python二维矩阵转置，不能使用numpy

# encoding=utf8


def show(L):
    # 直观显示用 非本题核心代码
    rows = len(L)
    cols = len(L[0])
    for i in range(rows):
        for j in range(cols):
            print(L[i][j], end=" ")
        print()

def transpose(A):
    B = []
    for i in range(len(A[0])):
        col = [0] * len(A)
        B.append(col)

    for i in range(len(A)):
        for j in range(len(A[0])):
            B[j][i] = A[i][j]
    print(‘转置前：‘)
    show(A)
    print(‘转置后：‘)
    show(B)


if __name__ == ‘__main__‘:
    A = [[1, 2, 3],
         [4, 5, 6]]
    transpose(A)

3.python计算两个bbox的iou 人家没说不能用numpy我就用了

# encoding=utf8
import numpy as np


def bbox_iou(gt_bbox, pred_bbox):
    ‘‘‘

    :param gt_bbox: 数据集bbox坐标
    :param pred_bbox: 预测bbox坐标
    :return: iou是否符合阈值  阈值默认0.5
    ‘‘‘
    # 得到union区域左上角和右下角
    insert_left = np.maximum(gt_bbox[:2], pred_bbox[:2])
    insert_right = np.minimum(gt_bbox[2:], pred_bbox[2:])
    # 重合区域面积
    insert_area = (insert_right[0] - insert_left[0]) * (insert_right[1] - insert_left[1])
    # gt_bbox面积
    gt_area = (gt_bbox[2] - gt_bbox[0]) * (gt_bbox[3] - gt_bbox[1])
    # pred_bbox面积
    pred_area = (pred_bbox[2] - pred_bbox[0]) * (pred_bbox[3] - pred_bbox[1])
    iou = insert_area / (gt_area + pred_area - insert_area)
    if iou > 0.5:
        return True
    else:
        return False


if __name__ == ‘__main__‘:
    # 例子
    gt_bbox   = [0.1, 0.1, 0.5, 0.5]
    pred_bbox = [0.3, 0.2, 0.7, 0.9]

    if bbox_iou(gt_bbox, pred_bbox):
        print(‘预测框达到IOU阈值‘)
    else:
        print(‘预测框未达到IOU阈值‘)

4.top_k查找 不允许使用sorted和list.sort

# encoding=utf8

def find_good_gay(file_name):
    ‘‘‘
    scores.txt文件在本项目文件夹中
    :param file_name: 文件路径
    :return: 成绩最好的三个学生的姓名和成绩
    ‘‘‘
    # 读取txt文件
    f = open(file_name)
    L = f.readlines()
    num = len(L)
    # 使用字典格式处理数据
    res_dict = {}

    # 生成成绩字典
    for i in range(num):
        res_dict[L[i].split(‘ ‘)[0]] = L[i].strip().split(‘ ‘)[1]
    # 排序
    for i in range(num):
        for j in range(num - i - 1):
            key1 = list(res_dict.keys())[j]
            key2 = list(res_dict.keys())[j + 1]
            if res_dict[key1] < res_dict[key2]:
                res_dict[key1], res_dict[key2] = res_dict[key2], res_dict[key1]
    print(‘成绩排在前三名的分别是：‘)
    for i in range(3):
        print(list(res_dict.keys())[i] + ‘:‘ + res_dict[list(res_dict.keys())[i]])


if __name__ == ‘__main__‘:
    find_good_gay(‘scores.txt‘)

随记-面试遇到的一些题

原文：https://www.cnblogs.com/haimingv/p/12631152.html